Chapter 2
Q. 2.18
Q. 2.18
The following data pertains to a tension test conducted in a strength of materials laboratory: diameter of the specimen = 15 mm, length of the specimen = 200 mm, extension under the laod of 10 kN = 0.035 mm, load at yield point = 110 kN, maximum load = 190 kN, length of specimen after failure = 255 mm, neck diameter = 12.25 mm. Determine: (i) Young’s modulus, (ii) yield stress, (iii) ultimate stress, (iv) percentage elongation, (v) percentage reduction in area, (vi) safe stress adopting a factor of safety of 1.5.
Step-by-Step
We calculate each of the parameters sought one by one.
Area of cross section of the specimen = A =\frac{\pi \times 20^2}{4}=314.16 mm ^2
(i) At load of 10 kN:
\text { stress }=\frac{P}{A}=\frac{10 \times 10^3}{314.16}=31.83 MPa
Strain corresponding to this load = \frac{0.035}{200}=1.75 \times 10^{-4}
∴ Modulus of elasticity E=\frac{31.83}{1.75 \times 10^{-4}}=1.81 \times 10^6 N / mm ^2
(ii) Stress at yield point:
∴ \sigma_{ y }=\frac{110 \times 10^3}{314.16}=350.14 N / mm ^2
(iii) Ultimate stress:
∴ \sigma_{ U }=\frac{190 \times 10^3}{314.16}=604.78 N / mm ^2
(iv) Percentage elongation:
\frac{255-200}{200} \times 100=27.5 \%
(v) Percentage reduction in area:
\frac{(\pi / 4)\left(15^2-12.25^2\right)}{(\pi / 4) 15^2} \times 100=33.3 \%